# Find a homogeneous linear differential equation with constant coefficients whose general solution is given by y=c1e−xcosx+c2e−xsinx.

Question

Find a homogeneous linear differential equation with constant coefficients whose general solution is given by
y=c1e−xcosx+c2e−xsinx.

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1 year 2021-09-04T15:15:49+00:00 1 Answers 16 views 0

## Answers ( )

1. Answer:

$$y” + 2y’ + 2y = 0$$

Step-by-step explanation:

Given

$$y=c_1e^{-x}cosx+c_2e^{-x}sinx$$

Required

Determine a homogeneous linear differential equation

Rewrite the expression as:

$$y=c_1e^{\alpha x}cos(\beta x)+c_2e^{\alpha x}sin(\beta x)$$

Where

$$\alpha = -1$$ and $$\beta = 1$$

For a homogeneous linear differential equation, the repeated value m is given as:

$$m = \alpha \± \beta i$$

Substitute values for $$\alpha$$ and $$\beta$$

$$m = -1 \± 1*i$$

$$m = -1 \± i$$

Add 1 to both sides

$$m +1= 1 -1 \± i$$

$$m +1= \± i$$

Square both sides

$$(m +1)^2= (\± i)^2$$

$$m^2 + m + m + 1 = i^2$$

$$m^2 + 2m + 1 = i^2$$

In complex numbers:

$$i^2 = -1$$

So, the expression becomes:

$$m^2 + 2m + 1 = -1$$

Add 1 to both sides

$$m^2 + 2m + 1 +1= -1+1$$

$$m^2 + 2m + 2= 0$$

This corresponds to the homogeneous linear differential equation

$$y” + 2y’ + 2y = 0$$